I have an impulse response of an LTI system which is

h(n)= n*u(n)-u(n-2) where n=[0,3] (1)

Now regarding the stability and causality of this system, i've drawn the conclusion that the system is causal since the output only relies on past or present values of n and it is stable since n is bounded. In this case n=0,1,2,3. Correct me here if i'm wrong but in case h(n) was u(n) - u(n-2) without knowing anything about n, it would stil be stable right? And if h(n) is the same as the original (1) but again without knowing anything about n it would be unstable correct?

Now i want to analyze this h(n) in odd and even components and also find the linear difference equation of h(n) ,but i can't find anything in my textbook that is of any help. It's the first time i've attended a signals and systems course and i'm really struggling to understand a few things, so any help here would be greatly appreciated.

  • $\begingroup$ Is $h[n]$ given by $n\cdot (u[n]-u[n-2])$ or by $n\cdot u[n] - u[n-2]$? $\endgroup$
    – Matt L.
    Commented Jul 7, 2020 at 16:06
  • $\begingroup$ @MattL. second one $\endgroup$ Commented Jul 7, 2020 at 16:22

1 Answer 1


You're right about your conclusions. And now a few hints to help you solve the questions:

  1. Your textbook should state somewhere the definition of even and odd parts of a (real-valued) sequence: $$h_e[n]=\frac12\big(h[n]+h[-n]\big)\\h_o[n]=\frac12\big(h[n]-h[-n]\big)$$

  2. In this case (FIR filter), the difference equation is simply given by the convolution of the input sequence $x[n]$ with the impulse response $h[n]$: $$y[n]=\sum_kx[k]h[n-k]$$

  • $\begingroup$ 1) Yes you are right. This was so early in the book(like page 9) i don't know how i missed it.Regarding (2) how do i know this is a FIR filter not a IIR one? Is it because i have discrete system, or because it's stable(meaning n is bounded)? $\endgroup$ Commented Jul 7, 2020 at 22:39
  • $\begingroup$ FIR stands for Finite impulse response. Since h(n) is not infinite it is FIR. It is only between 0 and 3 $\endgroup$
    – VMMF
    Commented Jul 8, 2020 at 0:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.